\newproblem{lay:6_3_2}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 6.3.2}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	You may assume that $\{\mathbf{u}_1,\mathbf{u}_2,\mathbf{u}_3,\mathbf{u}_4\}$ is an orthogonal basis of $\mathbb{R}^4$. Let
	$\mathbf{u}_1=(1,2,1,1)$, $\mathbf{u}_2=(-2,1,-1,1)$, $\mathbf{u}_3=(1,1,-2,-1)$, $\mathbf{u}_4=(-1,1,1,-2)$. Let $\mathbf{v}=(4,5,-3,3)$.
	Write $\mathbf{v}$ as the sum of two vectors, one in $\mathrm{Span}\{\mathbf{u}_1\}$ and the other in
	$\mathrm{Span}\{\mathbf{u}_2,\mathbf{u}_3,\mathbf{u}_4\}$.
}{
   % Solution
	We project $\mathbf{v}$ onto $\mathrm{Span}\{\mathbf{u}_1\}$ and then onto $\mathrm{Span}\{\mathbf{u}_{234}\}$:
	\begin{center}
		$\begin{array}{rcl}
			\mathbf{v}_{1}&=&\frac{\mathbf{v}\cdot\mathbf{u}_1}{\mathbf{u}_1\cdot\mathbf{u}_1}\mathbf{u}_1
											=\frac{14}{7}\begin{pmatrix}1\\2\\1\\1\end{pmatrix}
											 =\begin{pmatrix}2\\4\\2\\2\end{pmatrix}\\
			\mathbf{v}_{234}&=&\frac{\mathbf{v}\cdot\mathbf{u}_2}{\mathbf{u}_2\cdot\mathbf{u}_2}\mathbf{u}_2+
												 \frac{\mathbf{v}\cdot\mathbf{u}_3}{\mathbf{u}_3\cdot\mathbf{u}_3}\mathbf{u}_3+
												 \frac{\mathbf{v}\cdot\mathbf{u}_4}{\mathbf{u}_4\cdot\mathbf{u}_4}\mathbf{u}_4\\
											&=&\frac{3}{7}\begin{pmatrix}-2\\1\\-1\\1\end{pmatrix}+
												 \frac{12}{7}\begin{pmatrix}1\\1\\-2\\-1\end{pmatrix}+
												 \frac{-8}{7}\begin{pmatrix}-1\\1\\1\\-2\end{pmatrix}
											 =\begin{pmatrix}2\\1\\-5\\1\end{pmatrix}
		\end{array}$
	\end{center}
	It can be easily verified that $\mathbf{v}=\mathbf{v}_1+\mathbf{v}_{234}$.
}
\useproblem{lay:6_3_2}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
